437. Path Sum III

Medium

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /  \
    5   -3
   / \    \
  3   2   11
 / \   \
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

 

My answer is: basic idea to compute suffix sum of each path when node is being traversed. Thus complexity is O(n^2). Better solution is shown below, which is O(n).

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int pathFound = 0;
    private Map<TreeNode, Integer> nodesSum = new HashMap<TreeNode, Integer>();
    public int pathSum(TreeNode root, int sum) {
        
        List<List<Integer>> expectedValuesInPaths = new ArrayList<List<Integer>>();
        List<TreeNode> currentNodes = new ArrayList<TreeNode>();
        helper(root, sum, currentNodes);
        return pathFound;
    }
    
    private void helper(TreeNode node, int sum, List<TreeNode> currentNodesInPath) {
        
        if (node == null) {
            return;
        }
        int paths = getSuffixPaths(currentNodesInPath, node, sum);
        pathFound += paths;
        currentNodesInPath.add(node);
        helper(node.left, sum, currentNodesInPath);
        helper(node.right, sum, currentNodesInPath);
        // remove last added node
        currentNodesInPath.remove(currentNodesInPath.size() - 1);
    }
    
    private int getSuffixPaths(List<TreeNode> previousNodes, TreeNode currentNode, int sum) {
        int paths = 0;
        List<Integer> nodeSums = new ArrayList<Integer>();
        int currentSum = currentNode.val;
        // in case currentNode itself can be sufficient
        nodeSums.add(currentSum);
        if (currentSum == sum) {
                paths ++;
        }
        for (int i = previousNodes.size() - 1; i >= 0; i--) {
            TreeNode prevNode = previousNodes.get(i);
            currentSum += prevNode.val;
            nodeSums.add(currentSum);
            if (currentSum == sum) {
                paths ++;
            }
        }
        return paths;
    }
}

Best O(n) solution: calculate prefix sum for path. On 1 path, (prefix sum on index j - prefix sum on index i) = sum[i:j]. Thus, if prefixSum == targetValue OR (prefixSum - targetValue) in prefixSumsMap.

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public int pathSum(TreeNode root, int sum) { HashMap<Integer, Integer> preSum = new HashMap(); preSum.put(0,1); return helper(root, 0, sum, preSum); } public int helper(TreeNode root, int currSum, int target, HashMap<Integer, Integer> preSum) { if (root == null) { return 0; } currSum += root.val; int res = preSum.getOrDefault(currSum - target, 0); preSum.put(currSum, preSum.getOrDefault(currSum, 0) + 1); res += helper(root.left, currSum, target, preSum) + helper(root.right, currSum, target, preSum); // remove current node's impact preSum.put(currSum, preSum.get(currSum) - 1); return res; }